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Consider a surface that carries surface charge density. In electrostatics, boundary conditions are studied by showing that there is a discontinuity in the normal component of the electric field across the charged surface and that there is no discontinuity in the horizontal component.

My question is aren't the electric fields at the surface always perpendicular to the surface? I mean where does the horizontal component come from if wherever we look locally, close enough to the surface, it looks like a flat infinite patch and there is no horizontal components!

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Comments to the question(v1): What electrical properties are you assuming about the ambient material on both sides of the surface? Is your question essentially this question? – Qmechanic♦ Mar 14 '12 at 20:56
@Qmechanic Just vacuum on both sides of the interface. It is discussed in Griffith's in CH2 before talking about conductors or dielectrics or any of that, just charged interface. No, my question is not that question. – Revo Mar 15 '12 at 6:28

2 Answers

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As Qmechanic said, you have to consider the surrounding media. It seems like you are generalizing from conductors, which are not the only surfaces that can have a charge density!

As an example, consider the boundary of a dielectric with an applied external field $\vec{E}$. The component $\vec{E_{\perp}}$ will induce a surface charge and accompanying discontinuity in $\vec{E}$ by polarizing the medium. The field component parallel to the surface $\vec{E_{\parallel}}$ will be continuous across the surface and is not necessarily zero inside.

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No, no dielectrics or conductors, just a charged interface, with vacuum on both sides I suppose, if you have Griffiths please check CH2, the subsection on boundary conditions. – Revo Mar 15 '12 at 6:29
@Revo, there is no rule that E must be perpendicular to charged surfaces. I tried to illustrate this with a simple counterexample. – tmac Mar 15 '12 at 7:01
How about the classic example of finding the electric field of an infinite charged surface ? Using Gauss's law and applying symmetry principle, the electric field lines are parallel and perpendicular to the surface. – Revo Mar 15 '12 at 11:55
@Revo, In that case, the field will be perpendicular to the surface. There are plenty of cases you could cook up where the field is not perpendicular to the surface, as evidenced by the two examples given in answers here. – tmac Mar 15 '12 at 15:16
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Assume that both the surface and the bulk are insulators with vacuum permittivity $\varepsilon_0$, so that the charges cannot redistribute themselves.

  1. Consider first the electric field $$\vec{E}~=~\frac{\sigma}{2\varepsilon_0} \begin{pmatrix} {\rm sgn}(x)\\0\\0 \end{pmatrix} $$ associated with a uniformly charged capacitor plate at $x=0$ parallel to the $yz$ plane.

  2. Consider next the electric field $$\vec{E}~=~\frac{\sigma}{2\varepsilon_0} \begin{pmatrix} 0\\{\rm sgn}(y)\\0 \end{pmatrix} $$ associated with a uniformly charged capacitor plate at $y=0$ parallel to the $xz$ plane.

  3. Now construct a simple counterexample (where $\vec{E}$ is not perpendicular to the surface) by adding together the charge distributions in situation 1 and 2, cf. figure. Use superposition principle to determine $$\vec{E}~=~\frac{\sigma}{2\varepsilon_0} \begin{pmatrix} {\rm sgn}(x)\\{\rm sgn}(y)\\0 \end{pmatrix}, $$

Figure:

        Capacitor plate
              |
            \ | /         E-field lines
   \    \    \|/    /    / 
    \    \    |    /    /
     \    \   |   /    /
 -----------------------------  Capacitor plate
     /    /   |   \    \
    /    /    |    \    \ 
   /    /    /|\    \    \
            / | \
              |
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